Call Now to Set Up Tutoring:

(888) 888-0446

Linear Algebra : Linear Algebra

Study concepts, example questions & explanations for Linear Algebra

Create An Account

All Linear Algebra Resources

4 Diagnostic Tests 108 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #11 : Eigenvalues And Eigenvectors

$\begin{align*}&\text{Find any and all eigenvalues for the matrix }\\&A=\begin{bmatrix}-20&-7\\-14&12\end{bmatrix}\end{align*}$

Possible Answers:

$\lambda_{1}=-23.81;\lambda_{2}=13.81$

$\lambda_{1}=-29.81;\lambda_{2}=7.81$

$\lambda_{1}=-26.81;\lambda_{2}=10.81$

$\lambda_{1}=-22.81;\lambda_{2}=14.81$

Correct answer:

$\lambda_{1}=-22.81;\lambda_{2}=14.81$

Explanation:

$\begin{align*}&\text{Eigenvalues of a matrix, typically noted as }\lambda\text{, are those values which satisfy}\\&det(A-\lambda I)\text{(I being the identity matrix). For the matrix }A=\begin{bmatrix}-20&-7\\-14&12\end{bmatrix}\\&\text{We can represent that equation as follows }\begin{bmatrix}- \lambda - 20&-7\\-14&12 - \lambda\end{bmatrix}\\&\text{For a square matrix with dimensions of }2\\&det\begin{vmatrix} a&b \\ c&d \end{vmatrix}=ad-bc\\&\text{From that, we can find the eigenvalues:}\\&8\lambda + \lambda ^{2} - 338\\&\lambda_{1}=-22.81;\lambda_{2}=14.81\end{align*}$

Report an Error

Example Question #12 : Eigenvalues And Eigenvectors

$\begin{align*}&\text{Find any and all eigenvalues for the matrix }\\&A=\begin{bmatrix}-2&-17&-11\\17&-14&13\\2&20&-17\end{bmatrix}\end{align*}$

Possible Answers:

$\lambda_{1}=1.17+17.26i;\lambda_{2}=1.17-17.26i;\lambda_{3}=-29.33$

$\lambda_{1}=-9.83+17.26i;\lambda_{2}=-9.83-17.26i;\lambda_{3}=-40.33$

$\lambda_{1}=-0.83+17.26i;\lambda_{2}=-0.83-17.26i;\lambda_{3}=-31.33$

$\lambda_{1}=5.17+17.26i;\lambda_{2}=5.17-17.26i;\lambda_{3}=-25.33$

Correct answer:

$\lambda_{1}=-0.83+17.26i;\lambda_{2}=-0.83-17.26i;\lambda_{3}=-31.33$

Explanation:

$\begin{align*}&\text{Eigenvalues of a matrix, typically noted as }\lambda\text{, are those values which satisfy}\\&det(A-\lambda I)\text{(I being the identity matrix). For the matrix }A=\begin{bmatrix}-2&-17&-11\\17&-14&13\\2&20&-17\end{bmatrix}\\&\text{We can represent that equation as follows }\begin{bmatrix}- \lambda - 2&-17&-11\\17&- \lambda - 14&13\\2&20&- \lambda - 17\end{bmatrix}\\&\text{For a square matrix with dimensions of }3\\&det\begin{vmatrix} a&b&c \\ d&e&f\\g&h&i \end{vmatrix}=a(ei-fh)-b(di-fg)+c(dh-eg)\\&\text{From that, we can find the eigenvalues:}\\&(-1)\cdot (351\lambda + 33\lambda ^{2} + \lambda ^{3} + 9359)\\&\lambda_{1}=-0.83+17.26i;\lambda_{2}=-0.83-17.26i;\lambda_{3}=-31.33\end{align*}$

Report an Error

Example Question #13 : Eigenvalues And Eigenvectors

$\begin{align*}&\text{Calculate the eigenvalues for the matrix }\\&\begin{bmatrix}-4&-10\\12&-3\end{bmatrix}\end{align*}$

Possible Answers:

$\lambda_{1}=-0.50+10.94i;\lambda_{2}=-0.50-10.94i$

$\lambda_{1}=-3.50+10.94i;\lambda_{2}=-3.50-10.94i$

$\lambda_{1}=-10.50+10.94i;\lambda_{2}=-10.50-10.94i$

$\lambda_{1}=0.50+10.94i;\lambda_{2}=0.50-10.94i$

Correct answer:

$\lambda_{1}=-3.50+10.94i;\lambda_{2}=-3.50-10.94i$

Explanation:

$\begin{align*}&\text{Eigenvalues of a matrix A, often denoted as }\lambda\text{, are values which satisfy}\\&det(A-\lambda I)\text{, where I represents the identity matrix. For the given matrix }\begin{bmatrix}-4&-10\\12&-3\end{bmatrix}\\&\text{We can write a matrix of the form }\begin{bmatrix}- \lambda - 4&-10\\12&- \lambda - 3\end{bmatrix}\\&\text{For a square matrix with dimensions of }2\\&det\begin{vmatrix} a&b \\ c&d \end{vmatrix}=ad-bc\\&\text{Knowing this, we can expand and solve:}\\&7\lambda + \lambda ^{2} + 132\\&\lambda_{1}=-3.50+10.94i;\lambda_{2}=-3.50-10.94i\end{align*}$

Report an Error

Example Question #16 : Eigenvalues And Eigenvectors

$\begin{align*}&\text{Calculate the eigenvalues for the matrix }\\&\begin{bmatrix}-15&14&5\\-6&1&-4\\-17&-11&-15\end{bmatrix}\end{align*}$

Possible Answers:

$\lambda_{1}=-15.74+12.27i;\lambda_{2}=-15.74-12.27i;\lambda_{3}=2.49$

$\lambda_{1}=-24.74+12.27i;\lambda_{2}=-24.74-12.27i;\lambda_{3}=-6.51$

$\lambda_{1}=-14.74+12.27i;\lambda_{2}=-14.74-12.27i;\lambda_{3}=3.49$

$\lambda_{1}=-10.74+12.27i;\lambda_{2}=-10.74-12.27i;\lambda_{3}=7.49$

Correct answer:

$\lambda_{1}=-15.74+12.27i;\lambda_{2}=-15.74-12.27i;\lambda_{3}=2.49$

Explanation:

$\begin{align*}&\text{Eigenvalues of a matrix A, often denoted as }\lambda\text{, are values which satisfy}\\&det(A-\lambda I)\text{, where I represents the identity matrix. For the given matrix }\begin{bmatrix}-15&14&5\\-6&1&-4\\-17&-11&-15\end{bmatrix}\\&\text{We can write a matrix of the form }\begin{bmatrix}- \lambda - 15&14&5\\-6&1 - \lambda&-4\\-17&-11&- \lambda - 15\end{bmatrix}\\&\text{For a square matrix with dimensions of }3\\&det\begin{vmatrix} a&b&c \\ d&e&f\\g&h&i \end{vmatrix}=a(ei-fh)-b(di-fg)+c(dh-eg)\\&\text{Knowing this, we can expand and solve:}\\&(-1)\cdot (320\lambda + 29\lambda ^{2} + \lambda ^{3} - 992)\\&\lambda_{1}=-15.74+12.27i;\lambda_{2}=-15.74-12.27i;\lambda_{3}=2.49\end{align*}$

Report an Error

Example Question #14 : Eigenvalues And Eigenvectors

$\begin{align*}&\text{Find any and all eigenvalues for the matrix }\\&A=\begin{bmatrix}-7&16\\-5&-16\end{bmatrix}\end{align*}$

Possible Answers:

$\lambda_{1}=-7.50+7.73i;\lambda_{2}=-7.50-7.73i$

$\lambda_{1}=-11.50+7.73i;\lambda_{2}=-11.50-7.73i$

$\lambda_{1}=-4.50+7.73i;\lambda_{2}=-4.50-7.73i$

$\lambda_{1}=-8.50+7.73i;\lambda_{2}=-8.50-7.73i$

Correct answer:

$\lambda_{1}=-11.50+7.73i;\lambda_{2}=-11.50-7.73i$

Explanation:

$\begin{align*}&\text{Eigenvalues of a matrix, typically noted as }\lambda\text{, are those values which satisfy}\\&det(A-\lambda I)\text{(I being the identity matrix). For the matrix }A=\begin{bmatrix}-7&16\\-5&-16\end{bmatrix}\\&\text{We can represent that equation as follows }\begin{bmatrix}- \lambda - 7&16\\-5&- \lambda - 16\end{bmatrix}\\&\text{For a square matrix with dimensions of }2\\&det\begin{vmatrix} a&b \\ c&d \end{vmatrix}=ad-bc\\&\text{From that, we can find the eigenvalues:}\\&23\lambda + \lambda ^{2} + 192\\&\lambda_{1}=-11.50+7.73i;\lambda_{2}=-11.50-7.73i\end{align*}$

Report an Error

Example Question #15 : Eigenvalues And Eigenvectors

$\begin{align*}&\text{Find the eigenvalues for the matrix }\\&A=\begin{bmatrix}3&-18&-11\\-6&13&-20\\-19&-14&6\end{bmatrix}\end{align*}$

Possible Answers:

$\lambda_{1}=-25.30;\lambda_{2}=19.15+5.95i;\lambda_{3}=19.15-5.95i$

$\lambda_{1}=-22.30;\lambda_{2}=22.15+5.95i;\lambda_{3}=22.15-5.95i$

$\lambda_{1}=-27.30;\lambda_{2}=17.15+5.95i;\lambda_{3}=17.15-5.95i$

$\lambda_{1}=-29.30;\lambda_{2}=15.15+5.95i;\lambda_{3}=15.15-5.95i$

Correct answer:

$\lambda_{1}=-22.30;\lambda_{2}=22.15+5.95i;\lambda_{3}=22.15-5.95i$

Explanation:

$\begin{align*}&\text{Eigenvalues of a matrix A, usually denoted as }\lambda\text{, are values which satisfy}\\&det(A-\lambda I)\text{, where I is the identity matrix. For our matrix }A=\begin{bmatrix}3&-18&-11\\-6&13&-20\\-19&-14&6\end{bmatrix}\\&\text{We can define a matrix of the form }\begin{bmatrix}3 - \lambda&-18&-11\\-6&13 - \lambda&-20\\-19&-14&6 - \lambda\end{bmatrix}\\&\text{For a square matrix with dimensions of }3\\&det\begin{vmatrix} a&b&c \\ d&e&f\\g&h&i \end{vmatrix}=a(ei-fh)-b(di-fg)+c(dh-eg)\\&\text{Using this, we can solve for our eigenvalues:}\\&(-1)\cdot (\lambda ^{3} - 22\lambda ^{2} - 462\lambda + 11735)\\&\lambda_{1}=-22.30;\lambda_{2}=22.15+5.95i;\lambda_{3}=22.15-5.95i\end{align*}$

Report an Error

Example Question #16 : Eigenvalues And Eigenvectors

$\begin{align*}&\text{Calculate the eigenvalues for the matrix }\\&\begin{bmatrix}-13&-5&5\\11&-17&18\\11&-1&-3\end{bmatrix}\end{align*}$

Possible Answers:

$\lambda_{1}=-7.08+3.47i;\lambda_{2}=-7.08-3.47i;\lambda_{3}=-18.84$

$\lambda_{1}=-5.08+3.47i;\lambda_{2}=-5.08-3.47i;\lambda_{3}=-16.84$

$\lambda_{1}=-16.08+3.47i;\lambda_{2}=-16.08-3.47i;\lambda_{3}=-27.84$

$\lambda_{1}=-12.08+3.47i;\lambda_{2}=-12.08-3.47i;\lambda_{3}=-23.84$

Correct answer:

$\lambda_{1}=-7.08+3.47i;\lambda_{2}=-7.08-3.47i;\lambda_{3}=-18.84$

Explanation:

$\begin{align*}&\text{Eigenvalues of a matrix A, often denoted as }\lambda\text{, are values which satisfy}\\&det(A-\lambda I)\text{, where I represents the identity matrix. For the given matrix }\begin{bmatrix}-13&-5&5\\11&-17&18\\11&-1&-3\end{bmatrix}\\&\text{We can write a matrix of the form }\begin{bmatrix}- \lambda - 13&-5&5\\11&- \lambda - 17&18\\11&-1&- \lambda - 3\end{bmatrix}\\&\text{For a square matrix with dimensions of }3\\&det\begin{vmatrix} a&b&c \\ d&e&f\\g&h&i \end{vmatrix}=a(ei-fh)-b(di-fg)+c(dh-eg)\\&\text{Knowing this, we can expand and solve:}\\&(-1)\cdot (329\lambda + 33\lambda ^{2} + \lambda ^{3} + 1172)\\&\lambda_{1}=-7.08+3.47i;\lambda_{2}=-7.08-3.47i;\lambda_{3}=-18.84\end{align*}$

Report an Error

Example Question #20 : Eigenvalues And Eigenvectors

Give the characteristic polynomial of the matrix

$A = \begin{bmatrix} \frac{1}{2}& \frac{2}{3}\\ \\ -\frac{1}{4}& \frac{1}{3} \end{bmatrix}$

Possible Answers:

$\lambda ^{2} - \frac{5}{6} \lambda$

$\lambda ^{2} + \frac{5}{6} \lambda$

$\lambda ^{2} - \frac{5}{6} \lambda + \frac{1}{3}$

$\lambda ^{2} + \frac{5}{6} \lambda + \frac{1}{6}$

$\lambda ^{2} + \frac{5}{6} \lambda + \frac{1}{3}$

Correct answer:

$\lambda ^{2} - \frac{5}{6} \lambda + \frac{1}{3}$

Explanation:

The characteristic polynomial of a square matrix can be derived as follows:

Determine $\det( \lambda I - A)$ , using the identity matrix with the same dimensions as (two by two):

$\lambda I - A$

$= \lambda \begin{bmatrix} 1& 0\\ 0& 1 \end{bmatrix} - \begin{bmatrix} \frac{1}{2}& \frac{2}{3}\\ \\ -\frac{1}{4}& \frac{1}{3} \end{bmatrix}$

$= \begin{bmatrix} \lambda& 0\\ 0& \lambda \end{bmatrix} - \begin{bmatrix} \frac{1}{2}& \frac{2}{3}\\ \\ -\frac{1}{4}& \frac{1}{3} \end{bmatrix}$

Subtract the matrices by subtracting elementwise:

$=\begin{bmatrix} \lambda - \frac{1}{2}& 0- \frac{2}{3}\\ \\ 0 - \left (-\frac{1}{4} \right )& \lambda - \frac{1}{3} \end{bmatrix}$

$=\begin{bmatrix} \lambda - \frac{1}{2}& - \frac{2}{3}\\ \\ \frac{1}{4} & \lambda - \frac{1}{3} \end{bmatrix}$

Find the determinant of this matrix by taking the product of the upper left-to lower right diagonal and subtracting the product of the upper right-to-lower left diagonal:

$\det( \lambda I - A)$

$=\begin{vmatrix} \lambda - \frac{1}{2}& - \frac{2}{3}\\ \\ \frac{1}{4} & \lambda - \frac{1}{3} \end{vmatrix}$

$=\left ( \lambda - \frac{1}{2} \right )\left ( \lambda - \frac{1}{3} \right ) - \left ( - \frac{2}{3}\right ) \left ( \frac{1}{4} \right )$

$= \lambda ^{2} - \frac{5}{6} \lambda + \frac{1}{6} - \left ( - \frac{1}{6}\right )$

$= \lambda ^{2} - \frac{5}{6} \lambda + \frac{1}{3}$ ,

the correct choice.

Report an Error

Example Question #471 : Linear Algebra

True or false:

$\begin{bmatrix} 1\\ 2 \end{bmatrix}$ is an eigenvector of the matrix $\begin{bmatrix} 4& 1\\ 2& 5 \end{bmatrix}$ .

Possible Answers:

True

False

Correct answer:

True

Explanation:

A vector is an eigenvector of a matrix if and only if there exists a scalar value $\lambda$ - an eigenvalue - such that

$A v = \lambda v$ ;

or, equivalently, A v must be a scalar multiple of .

Letting $A = \begin{bmatrix} 4& 1\\ 2& 5 \end{bmatrix}$ and $v= \begin{bmatrix} 1\\ 2 \end{bmatrix}$ , find A v by multiplying each row of by - that is, multiplying each element in each row in by the corresponding element in . This is

$Av = \begin{bmatrix} 4& 1\\ 2& 5 \end{bmatrix}\begin{bmatrix} 1\\ 2 \end{bmatrix}$

$= \begin{bmatrix} 4(1)+ 1(2)\\ 2(1)+ 5(2) \end{bmatrix}$

$= \begin{bmatrix}4+2\\ 2+10 \end{bmatrix}$

$= \begin{bmatrix}6\\ 12 \end{bmatrix}$

Since $6 = 6 \cdot 1$ and $12 = 6 \cdot 2$ , it follows that

Av = 6v .

Therefore, such a $\lambda$ exists (and is equal to 6), and $v= \begin{bmatrix} 1\\ 2 \end{bmatrix}$ is indeed an eigenvector of $A = \begin{bmatrix} 4& 1\\ 2& 5 \end{bmatrix}$ .

Report an Error

Example Question #21 : Eigenvalues And Eigenvectors

True or false:

$\begin{bmatrix} 1\\ 1 \end{bmatrix}$ is an eigenvector of the matrix $\begin{bmatrix} 4& 1\\ 2& 5 \end{bmatrix}$ .

Possible Answers:

True

False

Correct answer:

False

Explanation:

A vector is an eigenvector of a matrix if and only if there exists a scalar value $\lambda$ - an eigenvalue - such that

$A v = \lambda v$ ;

or, equivalently, A v must be a scalar multiple of .

Letting $A = \begin{bmatrix} 4& 1\\ 2& 5 \end{bmatrix}$ and $v= \begin{bmatrix} 1\\ 1 \end{bmatrix}$ , find A v by multiplying each row of by - that is, multiplying each element in each row in by the corresponding element in . This is

$Av = \begin{bmatrix} 4& 1\\ 2& 5 \end{bmatrix}\begin{bmatrix} 1\\ 1 \end{bmatrix}$

$= \begin{bmatrix} 4(1)+ 1(1)\\ 2(1)+ 5(1) \end{bmatrix}$

$= \begin{bmatrix}4+1\\ 2+5\end{bmatrix}$

$= \begin{bmatrix}5\\ 7\end{bmatrix}$

$5 = 5 \cdot 1$ , but $7= 7 \cdot 1$ . Therefore, there cannot exist $\lambda$ such that $A v = \lambda v$ . This means that $v= \begin{bmatrix} 1\\ 1 \end{bmatrix}$ is not an eigenvector of $A = \begin{bmatrix} 4& 1\\ 2& 5 \end{bmatrix}$ .

Report an Error

View Linear Algebra Tutors

David
Certified Tutor

The University of Texas at El Paso, Bachelor of Science, Computational Mathematics. Indiana State University, Master of Scien...

View Linear Algebra Tutors

Ping
Certified Tutor

The University of Texas at Dallas, Bachelor of Science, Computer Science. University of North Texas, Master of Arts, Education.

View Linear Algebra Tutors

Shiva Tej
Certified Tutor

Vellore Institute of Technology, Bachelor of Science, Mechanical Engineering. University of Toronto, Master of Science, Data ...

All Linear Algebra Resources

4 Diagnostic Tests 108 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

Reading Tutors in Atlanta, Statistics Tutors in Miami, Physics Tutors in Los Angeles, GRE Tutors in Washington DC, LSAT Tutors in Washington DC, GMAT Tutors in New York City, Biology Tutors in Seattle, SSAT Tutors in Denver, Calculus Tutors in Chicago, SSAT Tutors in Los Angeles

Popular Courses & Classes

Spanish Courses & Classes in New York City, ACT Courses & Classes in Los Angeles, ISEE Courses & Classes in Phoenix, GRE Courses & Classes in Seattle, SSAT Courses & Classes in Dallas Fort Worth, SSAT Courses & Classes in Houston, GRE Courses & Classes in Boston, ISEE Courses & Classes in Chicago, Spanish Courses & Classes in Washington DC, ACT Courses & Classes in Miami

Popular Test Prep

SAT Test Prep in Atlanta, SSAT Test Prep in Boston, ACT Test Prep in San Diego, ACT Test Prep in Philadelphia, GRE Test Prep in San Diego, SAT Test Prep in Washington DC, LSAT Test Prep in Houston, ACT Test Prep in Los Angeles, ISEE Test Prep in Phoenix, SAT Test Prep in San Francisco-Bay Area

DMCA Complaint

If you believe that content available by means of the Website (as defined in our Terms of Service) infringes one or more of your copyrights, please notify us by providing a written notice (“Infringement Notice”) containing the information described below to the designated agent listed below. If Varsity Tutors takes action in response to an Infringement Notice, it will make a good faith attempt to contact the party that made such content available by means of the most recent email address, if any, provided by such party to Varsity Tutors.

Your Infringement Notice may be forwarded to the party that made the content available or to third parties such as ChillingEffects.org.

Please be advised that you will be liable for damages (including costs and attorneys’ fees) if you materially misrepresent that a product or activity is infringing your copyrights. Thus, if you are not sure content located on or linked-to by the Website infringes your copyright, you should consider first contacting an attorney.

Please follow these steps to file a notice:

You must include the following:

A physical or electronic signature of the copyright owner or a person authorized to act on their behalf; An identification of the copyright claimed to have been infringed; A description of the nature and exact location of the content that you claim to infringe your copyright, in \ sufficient detail to permit Varsity Tutors to find and positively identify that content; for example we require a link to the specific question (not just the name of the question) that contains the content and a description of which specific portion of the question – an image, a link, the text, etc – your complaint refers to; Your name, address, telephone number and email address; and A statement by you: (a) that you believe in good faith that the use of the content that you claim to infringe your copyright is not authorized by law, or by the copyright owner or such owner’s agent; (b) that all of the information contained in your Infringement Notice is accurate, and (c) under penalty of perjury, that you are either the copyright owner or a person authorized to act on their behalf.

Send your complaint to our designated agent at:

Charles Cohn Varsity Tutors LLC
101 S. Hanley Rd, Suite 300
St. Louis, MO 63105

Or fill out the form below:

Do not fill in this field

Contact Information

* Full legal name

Company name

* Phone number

* Email address

Address

* Address1

Address2

* City

* State

* Zipcode

* Country

Complaint Details

* URL of copyrighted work (where your original material is located)

* Please describe the copyrighted work so that it may be easily identified

I am the owner, or an agent authorized to act on behalf of the owner of an exclusive right that is allegedly infringed.

I have a good faith belief that the use of the material in the manner complained of is not authorized by the copyright owner, its agent, or the law.

This notification is accurate.

I acknowledge that there may be adverse legal consequences for making false or bad faith allegations of copyright infringement by using this process.

* Signature (your digital signature is legally binding)

HIGH SCHOOL

GRADUATE SCHOOL

K-8

Search 50+ Tests

math tutoring

science tutoring

foreign languages

elementary tutoring

other

Search 350+ Subjects

Linear Algebra : Linear Algebra

Study concepts, example questions & explanations for Linear Algebra

All Linear Algebra Resources

Example Questions

Example Question #11 : Eigenvalues And Eigenvectors

Example Question #12 : Eigenvalues And Eigenvectors

Example Question #13 : Eigenvalues And Eigenvectors

Example Question #16 : Eigenvalues And Eigenvectors

Example Question #14 : Eigenvalues And Eigenvectors

Example Question #15 : Eigenvalues And Eigenvectors

Example Question #16 : Eigenvalues And Eigenvectors

Example Question #20 : Eigenvalues And Eigenvectors

Example Question #471 : Linear Algebra

Example Question #21 : Eigenvalues And Eigenvectors

All Linear Algebra Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: